A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is t a n - 1 1 2 .…

The given water tank is of the shape shown by the diagram.

The semi-vertical angle $\theta ={\tan }^{-1}\left(\frac{1}{2}\right), \Rightarrow \tan \theta =\frac{1}{2}$

Let at any time $t min,$ height of water level is $h m$  and radius of cone filled with water be $r m.$

Also, we have $\tan \theta =\frac{r}{h}$

$\Rightarrow \frac{r}{h}=\frac{1}{2}$

$\Rightarrow r=\frac{h}{2} ...\left(i\right)$

Now, the volume of the water at time $t min$ in the cone is $V=\frac{1}{3}\pi r^{2}h$

On putting the value of $r$ from the equation $\left(i\right),$ we get

$V=\frac{\mathrm{\pi }}{3}\left(\frac{h^3}{4}\right)=\frac{\mathrm{\pi }}{12}\left(h^3\right)$

Now, differentiating with respect to $t,$ we get

$\frac{dV}{dt}=\frac{\pi }{12}\cdot \left(3h^2\frac{dh}{dt}\right)$

Put the given value of $\frac{dV}{dt} \& h$

$\Rightarrow 5=\frac{\pi }{4}\cdot \left(100\frac{dh}{dt}\right)$

$\Rightarrow \frac{dh}{dt}=\frac{1}{5\mathrm{\pi }} m/min$